Problem: Multiply the following complex numbers: $({-1+4i}) \cdot ({4-3i})$
Explanation: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-1+4i}) \cdot ({4-3i}) = $ $ ({-1} \cdot {4}) + ({-1} \cdot {-3}i) + ({4}i \cdot {4}) + ({4}i \cdot {-3}i) $ Then simplify the terms: $ (-4) + (3i) + (16i) + (-12 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -4 + (3 + 16)i - 12i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -4 + (3 + 16)i - (-12) $ The result is simplified: $ (-4 + 12) + (19i) = 8+19i $